Question

Given two points P(sinθ+2, tanθ-2) and Q(4sin²θ+4sinθcosθ+2acosθ, 3sinθ-2cosθ+a). Find constant "a" and the corresponding value of θ when these two points coincide. (0 ≤ θ < 2π)

Show your work, thanks!​

Answer 1

`\rm\displaystyle \displaystyle \displaystyle θ= {60}^( \circ) , {300}^( \circ)`

`\rm \displaystyle a = - ( √(3) )/(2) - 1, (√(3))/(2) - 1`

Explanation:

we are given two coincident points

`\displaystyle P( \sin(θ)+2, \tan(θ)-2) \: \text{and } \\ \displaystyle Q(4 \sin ^(2) (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)`

since they are coincident points

`\rm \displaystyle P( \sin(θ)+2, \tan(θ)-2) = \displaystyle Q(4 \sin ^(2) (θ)+4 \sin(θ )\cos(θ)+2a \cos(θ), 3 \sin(θ)-2 \cos(θ)+a)`

By order pair we obtain:

`\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^(2) (θ)+4 \sin(θ) \cos(θ)+2a \cos(θ) = \sin( \theta) + 2 \\ \\ \displaystyle 3 \sin( \theta) - 2 \cos( \theta) + a = \tan( \theta) - 2\end{cases}`

now we end up with a simultaneous equation as we have two variables

to figure out the simultaneous equation we can consider using substitution method

to do so, make a the subject of the equation.therefore from the second equation we acquire:

`\begin{cases} \rm\displaystyle \displaystyle 4 \sin ^(2) (θ)+4 \sinθ \cos(θ)+2a \cos(θ )= \sin( \theta) + 2 \\ \\ \boxed{\displaystyle a = \tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) } \end{cases}`

now substitute:

`\rm\displaystyle \displaystyle 4 \sin ^(2) (θ)+4 \sin(θ) \cos(θ)+2 \cos(θ) \{\tan( \theta) - 2 - 3 \sin( \theta) + 2 \cos( \theta) \}= \sin( \theta) + 2`

distribute:

`\rm\displaystyle \displaystyle 4 \sin ^(2)( θ)+4 \sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) - 6 \sin( \theta) \cos( \theta) + 4 \cos ^(2) ( \theta) = \sin( \theta) + 2`

collect like terms:

`\rm\displaystyle \displaystyle 4 \sin ^(2)( θ) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + 4 \cos ^(2) ( \theta) = \sin( \theta) + 2`

rearrange:

`\rm\displaystyle \displaystyle 4 \sin ^(2)( θ) + 4 \cos ^(2) ( \theta) - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) + = \sin( \theta) + 2`

by Pythagorean theorem we obtain:

`\rm\displaystyle \displaystyle 4 - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) + 2`

cancel 4 from both sides:

`\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+2 \sin(θ ) - 4\cos( \theta) = \sin( \theta) - 2`

move right hand side expression to left hand side and change its sign:

`\rm\displaystyle \displaystyle - 2\sin(θ) \cos(θ)+\sin(θ ) - 4\cos( \theta) + 2 = 0`

factor out sin:

`\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) - 4\cos( \theta) + 2 = 0`

factor out 2:

`\rm\displaystyle \displaystyle \sin (θ) (- 2 \cos(θ)+1) + 2(- 2\cos( \theta) + 1 ) = 0`

group:

`\rm\displaystyle \displaystyle ( \sin (θ) + 2)(- 2 \cos(θ)+1) = 0`

factor out -1:

`\rm\displaystyle \displaystyle - ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0`

divide both sides by -1:

`\rm\displaystyle \displaystyle ( \sin (θ) + 2)(2 \cos(θ) - 1) = 0`

by Zero product property we acquire:

`\begin{cases}\rm\displaystyle \displaystyle \sin (θ) + 2 = 0 \\ \displaystyle2 \cos(θ) - 1= 0 \end{cases}`

cancel 2 from the first equation and add 1 to the second equation since -1≤sinθ≤1 the first equation is false for any value of theta

`\begin{cases}\rm\displaystyle \displaystyle \sin (θ) \\eq - 2 \\ \displaystyle2 \cos(θ) = 1\end{cases}`

divide both sides by 2:

`\rm\displaystyle \displaystyle \displaystyle \cos(θ) = (1)/(2)`

by unit circle we get:

`\rm\displaystyle \displaystyle \displaystyle θ= {60}^( \circ) , {300}^( \circ)`

so when θ is 60° a is:

`\rm \displaystyle a = \tan( {60}^( \circ) ) - 2 - 3 \sin( {60}^( \circ) ) + 2 \cos( {60}^( \circ) )`

recall unit circle:

`\rm \displaystyle a = √(3) - 2 - ( 3√(3) )/(2) + 2 \cdot (1)/(2)`

simplify which yields:

`\rm \displaystyle a = - ( √(3) )/(2) - 1`

when θ is 300°

`\rm \displaystyle a = \tan( {300}^( \circ) ) - 2 - 3 \sin( {300}^( \circ) ) + 2 \cos( {300}^( \circ) )`

remember unit circle:

`\rm \displaystyle a = - √(3) - 2 + (3√( 3) )/(2) + 2 \cdot (1)/(2)`

simplify which yields:

`\rm \displaystyle a = ( √(3) )/(2) - 1`

and we are done!

disclaimer: also refer the attachment I did it first before answering the question